2.6 WORKED EXAMPLES
1) A metallic bar 300 mm x 100 mm x 40 mm is subjected to a force of 5 kN (tensile), 6 kN
(tensile) and 4 kN (tensile) along x, y and z directions respectively. Determine the change in
the volume of the block. Take E = 2 x 105 N/mm2 and Poisson's ratio = 0.25.
Solution:
Given: Dimensions of bar
= 300 mm x 100 mm x 40 mm
x = 300 mm, y = 100 mm and z = 40 mm
V= xyz = 300 x 100x40 = 1200000 mm3
Load in the direction of x=5KN; Load in the direction of y =6KN; Load in the direction of z =
4KN
2) A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge
length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate: (i) Young's
modulus, (ii) Poisson's ratio and (iii) Bulk modulus.
3) A rod is 2 m long at a temperature of 10°C. Find the expansion of the rod, when the
temperature is raised to 80°C. If this expansion is prevented, find the stress induced in the
material of the rod. Take E=1.0x 105 MN/m2 and  = 0.000012 per degree centigrade.
4) A steel rod of 20 mm diameter passes centrally through a copper tube of 50 mm external
diameter and 40 mm internal diameter. The tube is closed at each end by rigid plates of
negligible thickness. The nuts are tightened lightly home on the projecting parts of the rod. If
the temperature of the assembly is raised by 50°C, calculate the stresses developed in copper
and steel. Take E for steel and copper as 200 GN/m2 and 100 GN/m2 and  for steel and
copper as 12 x 10-6 per °C and 18 x 10-6 per °C.
As  for copper is more than that of steel, hence the free expansion of copper will be more than
that of steel when there is a rise in temperature. But the ends of the rod and the tube is fixed to
the rigid plates and the nuts are tightened on the projected parts of the rod. Hence the two
members are not free to expand. Hence the tube and the rod will expand by the same amount.
The free expansion of the copper tube will be more than the common expansion, whereas the free
expansion of the steel rod will be less than the common expansion. Hence the copper tube will be
subjected to compressive stress and the steel rod will be subjected to tensile stress.
Let
s = Tensile stress in steel

c = Compressive stress in copper.
For the equilibrium of the system,
Tensile load on steel = Compressive load on copper
…(i)
We know that the copper tube and steel rod will actually expand by the same amount. Actual
expansion of steel = Actual expansion of copper
… (ii)
But actual expansion of steel
= Free expansion of steel + Expansion due to tensile stress in steel
and actual expansion of copper
= Free expansion of copper - Contraction due to compressive stress in copper
Substituting these values in equation (ii), we get